Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int arr[100050];
void tr() {
  int i;
  for (i = 1; i <= 100000; i++) {
    arr[i] = arr[i - 1] + i;
  }
}
bool binary(long long int arr[], int n, long long int search_element) {
  long long int high, low, mid;
  low = 1;
  high = n + 1;
  mid = (low + high) / 2;
  while (low < high) {
    if (arr[mid] == search_element)
      return true;
    else if (search_element < arr[mid]) {
      high = mid;
      mid = (high + low) / 2;
    } else if (search_element > arr[mid]) {
      low = mid + 1;
      mid = (high + low) / 2;
    }
  }
  return false;
}
int main() {
  long long int n, i, j, k, result;
  tr();
  scanf("%I64d", &n);
  for (i = 1; i < (100000 / 2); i++) {
    k = n - arr[i];
    result = binary(arr, 100000, k);
    if (result == true) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}